2 Answers
2

For example, if you substitute the third equation into the first, you get that

$$ [1-\alpha(1-p)](\pi_1-\pi_4)=0$$

For now, I'm going to assume that the coefficient is non-zero, which gives $\pi_1 = \pi_4$. The second and third equation then give that $\pi_2=\pi_3$. Substituting this into the last equation, we get that $\pi_1 = \frac {1}{ 2 +2\alpha(1-p)}$ (which is equal to $\pi_4$). And the value of $\pi_3 = \pi_2$ drops out immediately.

Now, if the coefficient is 0, then $[1-\alpha(1-p)]=0$, which gives $\pi_1=\pi_3$, $\pi_2=\pi_4$, and you have infinitely many solutions subject to $\pi_1+\pi_2 = \frac {1}{2}$.

You should deal with the $w=1$ case separately, since $w\neq 1$ wasn't mentioned, and this leads to additional answers. In particular, if OP is using $w=1$, then they will get all the $\pi_i$ being the same, which was their initial problem.
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Calvin LinJan 19 '13 at 23:06

@CalvinLin: Thanks for the comment. From (6), I would have thought that would be clear, but all I am showing is the π1,π2,π3,π4 issue the OP is asking about. Regards
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AmzotiJan 19 '13 at 23:17